Exercise 6.4.3: Glare from a windscreen

$\sin {\theta }_t=\sin 70^{\circ }/1.5=0.626$, ${\theta }_t=38.8^{\circ }$, $\cos {\theta }_i=0.342$, $\cos {\theta }_t=0.779$. With ${\eta }_1=1$, ${\eta }_2=1/1.5$:

$R_{\perp }=0.300$, $R_{\parallel }=0.042$, and for sunlight $R=\frac{1}{2}(0.300+0.042)=0.171$: four times the $4\%$ of normal incidence.

$P=\frac{0.300-0.042}{0.300+0.042}=0.75$: the glare is $75\%$ polarised, with $E$ mostly parallel to the glass (S).

$\frac{R_{\parallel }}{R_{\perp }+R_{\parallel }}=\frac{0.042}{0.342}=0.124$: the glasses cut the glare to $12\%$, by $10{\log }_{10}(1/0.124)=9.1\mathrm{dB}$.

At the Brewster angle ${\theta }_B=\arctan 1.5=56.3^{\circ }$ the reflected light is pure S and is blocked completely.

The angle forms of Home exercise 6.4.2 give ${\Gamma }_{\perp }=-\sin 31.2^{\circ }/\sin 108.8^{\circ }=-0.547$ and ${\Gamma }_{\parallel }=-\tan 31.2^{\circ }/\tan 108.8^{\circ }=+0.206$. ${\Gamma }_{\parallel }$ is negative at normal incidence ($-0.2$), passes through zero at ${\theta }_B$ and is positive beyond it.